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3 years ago

Triangle DEF has vertices D(0,4), E(10,4), and F(5,9) Prove that the triangle is isosceles and a right triangle.

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

8 0

Answer:

darling, could you send a picture of the problem?

Step-by-step explanation:

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8.75 is 0.5% of what number

Otrada [13]

Answer:

1750

Step-by-step explanation:

I put it in an image

but multiplyng it would have been the opposite.

and it would be 0.5% of 8.75 is what.

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3 years ago

What causes a repeating decimal in the long division algorithm?

Nady [450]

Answer:

Step-by-step explanation:

In the division algorithm, if the remainder is zero, then the algorithm terminates, resulting in a terminating decimal. remainder repeats, the calculations that follow will also repeat in a cyclical pattern causing a repeating decimal.

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3 years ago

In square YOLA, RL=5x+28 and RO=8X-11

Alexeev081 [22]

Answer:

RL=5x+28 and

RO=8X-11

diagonal of square bisect equally the side

:.5x+28=8x-11

11+28=8x-5x

39=3x

x=39/3=13

RY=RL=5x+28=5×13+28=93If the answer is 93, move to answer.

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2 years ago

Mal is putting money into a savings account. She starts with $350 in the savings account, and each week she adds $30.

Bogdan [553]

S=350+(30*w)
so
350+(30*19)=920

4 0

2 years ago

Read 2 more answers

Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

Sergio [31]

Answer:

$\bold{cot(t) =\dfrac{12}{5}}$

Step-by-step explanation:

Given that:

$Cosec (t) = -\frac{13}5$

for $\pi < t < \frac{3 \pi}2$

That means, angle $t$ is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

$cosec^2\theta -cot^2\theta =1$

here, $\theta =t$

So, $cosec^2t-cot^2t=1$

$\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}$

But, angle $t$ is in 3rd quadrant, so value of

$\bold{cot(t) =\dfrac{12}{5}}$

4 0

3 years ago